# Finite non-solvable groups whose real degrees are prime-powers

Lorenzo Bonazzi
From the journal Journal of Group Theory

# Abstract

We present a description of non-solvable groups in which all real irreducible character degrees are prime-power numbers.

## 1 Introduction

Let 𝐺 be a finite group. It is well known that cd ( G ) , the set of the degrees of all irreducible characters, has great impact on the structure of 𝐺. Manz in [12, 11] described the solvable and non-solvable groups in which all the real irreducible characters have prime-power degrees. In this paper, we give a structural description of non-solvable groups 𝐺 such that cd rv ( G ) , the set of the degrees of all real irreducible characters, consists of prime-power numbers. In the following, Rad ( G ) is the radical subgroup and G ( ) is the last term of the derived series.

Theorem A

Let 𝐺 be a finite non-solvable group, and suppose that cd rv ( G ) consists of prime-power numbers. Then Rad ( G ) = H × O for a group 𝑂 of odd order and a 2-group 𝐻 of Chillag–Mann type. Furthermore, if K = G ( ) , then one of the following holds:

1. G = K × Rad ( G ) , and 𝐾 is isomorphic to A 5 or L 2 ( 8 ) ;

2. G = ( K H ) × O , where K SL 2 ( 5 ) , H K = H \Ydown K , K H = Z ( K ) < H .

About (ii), we remark that if 𝐺 is the SmallGroup(240,93), then K SL 2 ( 5 ) , | H | = 4 and K H = Z ( K ) .

As a corollary, we get control on the set of real character degrees. We recall that cd rv , 2 ( G ) is the set of odd real character degrees of a finite group 𝐺.

Theorem B

Let 𝐺 a non-solvable group such that cd rv ( G ) consists of prime-power numbers. Then either

1. cd rv ( G ) = cd rv ( L 2 ( 8 ) ) or

2. cd rv , 2 ( G ) = cd rv , 2 ( A 5 ) .

## 2 Preliminary results and lemmas

Chillag and Mann are among the first authors that studied cd rv ( G ) . They characterized the groups 𝐺 such that cd rv ( G ) = { 1 } , namely where all real irreducible characters are linear. Now these groups are commonly known as groups of Chillag–Mann type.

Theorem 2.1

## Theorem 2.1 ([2, Theorem 1.1])

Let 𝐺 be a finite group of Chillag–Mann type. Then G = O × T , where 𝑂 is a group of odd order and 𝑇 is a 2-group of Chillag–Mann type.

One other important contribution was given by Dolfi, Navarro and Tiep in [3]. In that paper, there appears a version for real characters of the celebrated Ito–Michler Theorem for the prime p = 2 . Recall that Irr rv ( G ) denotes the set of irreducible real-valued character of 𝐺.

Theorem 2.2

## Theorem 2.2 ([3, Theorem A])

Let 𝐺 be a finite group and T Syl 2 ( G ) . Then 2 χ ( 1 ) for every non-linear χ Irr rv ( G ) if and only if T G and 𝑇 is of Chillag–Mann type.

The corresponding condition for an odd prime 𝑝 was studied by Tiep in [15] and Isaacs and Navarro in [9]. Though a partial result, the techniques involved are deep. This confirms the special role of the prime 2 in the study of real character degrees.

Theorem 2.3

## Theorem 2.3 ([15, Theorem A])

Let 𝐺 be a finite group and 𝑝 a prime. Suppose that p χ ( 1 ) for every χ Irr rv ( G ) with Schur–Frobenius indicator 1. Then O p ( G ) is solvable; in particular, 𝐺 is 𝑝-solvable.

Navarro, Sanus and Tiep gave a version for real characters of Thompson’s Theorem for the prime 2 in [13]. Their work includes also a characterization of groups in which the real character degrees are powers of 2.

Theorem 2.4

## Theorem 2.4 ([13, Theorem A])

Let 𝐺 be a finite group, and suppose that 2 divides χ ( 1 ) for all every real non-linear irreducible character of 𝐺. Then 𝐺 has a normal 2-complement.

The next two lemmas appear in [4].

Lemma 2.5

Let 𝑁 be a normal subgroup of 𝐺 and χ Irr rv ( G ) . The following hold:

1. if χ ( 1 ) is odd, then N ker ( χ ) ;

2. if | N | is odd and 𝑁 centralizes a Sylow 2-subgroup of 𝐺, then N ker ( χ ) .

## Proof

Point (ii) is [4, Lemma 1.4]. Point (i) follows from the discussion before [4, Lemma 1.4], keeping in mind that a group of odd order does not have any real non-trivial character. ∎

Let 𝑁 be a normal subgroup of 𝐺 and θ Irr rv ( N ) . The next lemmas provide some sufficient conditions for the existence of a real character of 𝐺 above 𝜃.

Lemma 2.6

## Lemma 2.6 ([14, Lemma 2.1 and Corollary 2.2])

Let 𝑁 be a normal subgroup of a group 𝐺 and θ Irr rv ( N ) . If [ G : N ] is odd, then 𝜃 allows a unique real-valued extension to I G ( θ ) . Furthermore, there exists a unique real-valued character χ Irr rv ( G θ ) .

Lemma 2.7

## Lemma 2.7 ([14, Thorem 2.3])

Let 𝐺 be a finite group and N G . Suppose that there is some θ Irr rv ( G ) such that θ ( 1 ) is odd and o ( θ ) = 1 . Then 𝜃 extends to a character ϕ Irr rv ( I G ( θ ) ) and χ = ϕ G Irr rv ( G θ ) .

Lemma 2.8

Suppose that 𝑁 is a minimal normal subgroup of a group 𝐺 with N = S 1 × × S n , where S S is a non-abelian simple group. Let σ Irr rv ( S ) , and suppose that 𝜎 extends to a real character of Aut ( S ) . Then σ × × σ extends to a real character of 𝐺.

## Proof

The extension 𝜒 is constructed in [1, Lemma 5]. We see that if 𝜎 takes real values, then so does 𝜒. ∎

The technique used in the proof of Lemma 2.8 is known as tensor induction; for further details, see [7, Section 4].

Lemma 2.9

## Lemma 2.9 ([4, Lemma 1.6])

Let 𝐺 be a finite group that acts by automorphisms on the group 𝑀. For every involution x C G ( M ) G / C G ( M ) , there exists a non-trivial character μ Irr ( M ) such that μ x = μ ¯ .

## 3 Proofs

In the following, we call an integer a composite number if it is divisible by more than one prime. If 𝑝 is a prime, we denote by p * a general positive integer that is a power of 𝑝. Moreover, Rad ( G ) is the solvable radical of 𝐺, namely the largest solvable normal subgroup of 𝐺.

Theorem 3.1

Let 𝐺 be a finite non-solvable group such that cd rv ( G ) consists of prime-power numbers. If Rad ( G ) = 1 , then 𝐺 is isomorphic to A 5 or PSL 2 ( 8 ) .

## Proof

Let 𝑀 be a minimal normal subgroup of 𝐺. Then M = S 1 × × S n is the product of simple groups, which are all isomorphic to a simple group 𝑆. Since Rad ( G ) = 1 , the group 𝑆 is non-abelian.

Step 1: 𝑆 is isomorphic to one of the following groups:

A 5 , A 6 , PSL 2 ( 8 ) , PSL 3 ( 3 ) , PSp 4 ( 3 ) , PSL 2 ( 7 ) , PSU 3 ( 3 ) , PSL 2 ( 17 ) .

Let p π ( M ) . Since 𝑀 is minimal normal in 𝐺, we have M O p ( G ) , so O p ( G ) is non-solvable. By Theorem 2.3, there is a real irreducible character 𝜒 of 𝐺 such that p χ ( 1 ) . By the hypothesis, χ ( 1 ) = p * > 1 . This means that, for every prime p π ( M ) , there is χ Irr rv ( G ) such that χ ( 1 ) = p * > 1 . By [3, Theorem B], if Δ rv ( G ) is the prime graph on real character degrees of 𝐺, then the number of connected components of Δ rv ( G ) is at most three. In our hypotheses, Δ rv ( G ) consists of isolated vertices, and hence the number of primes that appear as divisors of the degree of some real irreducible character is at most 3. It follows that 𝑀, and hence 𝑆, is divisible by exactly 3 primes. Now, by [16, Lemma 2.1], the simple groups having order divided by exactly 3 distinct primes are those stated.

Step 2: 𝑆 is isomorphic to one of the following groups: A 5 , PSL 2 ( 8 ) , A 6 .

If S { PSp 4 ( 3 ) , PSL 3 ( 3 ) , PSU 3 ( 3 ) } , there is a non-linear character σ Irr rv ( S ) such that σ ( 1 ) is an odd composite number. Let θ = σ × × σ Irr rv ( M ) . Then we have 2 θ ( 1 ) and o ( θ ) = 1 since 𝑀 is perfect. So, by Lemma 2.7, there is χ Irr rv ( G θ ) . As θ ( 1 ) divides χ ( 1 ) , the degree of 𝜒 is a composite number, contrary to the hypothesis. Suppose that S { PSL 2 ( 7 ) , PSL 2 ( 17 ) } . There is a real character σ Irr rv ( S ) such that σ ( 1 ) is a composite number and 𝜎 extends to a real character of A = Aut ( S ) . By tensor induction (Lemma 2.8), the character θ = σ × × σ extends to a real character χ Irr rv ( G ) . Again, we get that χ ( 1 ) = θ ( 1 ) = σ ( 1 ) n is a composite number.

Step 3: n = 1 and 𝑀 is a simple group.

The only remaining possibilities are S { A 5 , PSL 2 ( 8 ) , A 6 } . Checking the character table of these groups, there are two non-linear characters σ , ρ Irr rv ( S ) such that σ ( 1 ) = p * > 1 and ρ ( 1 ) = q * > 1 for p , q odd distinct primes. Let θ = σ × 1 × × 1 Irr rv ( M ) . Since o ( θ ) = 1 and θ ( 1 ) is odd, the character 𝜃 extends to a character ϕ Irr rv ( I G ( θ ) ) by Lemma 2.7, and χ = ϕ G has degree p * ; hence [ G : I G ( θ ) ] = p * > 1 . Since I G ( θ ) N G ( S 1 ) , we have that

n = [ G : N G ( S 1 ) ] divides [ G : I G ( θ ) ] = p * > 1 ,

so n = p * . By the same argument with 𝜌 in place of 𝜎, we get that n = q * and n ( p * , q * ) = 1 .

Step 4: C G ( M ) = 1 .

Suppose, by contradiction, that C G ( M ) > 1 , and take 𝑁 to be a minimal normal subgroup of 𝐺 contained in C G ( M ) . By the same arguments as used on 𝑀, we have that 𝑁 is isomorphic to a simple group of the following: A 5 , PSL 2 ( 8 ) , A 6 . As before, take σ Irr rv ( M ) with σ ( 1 ) = p * and ρ Irr rv ( N ) with ρ ( 1 ) = q * for p , q odd distinct primes. Note that [ M , N ] M N M C G ( M ) = 1 since 𝑀 is simple and non-abelian. So M N = M × N is perfect normal in 𝐺, and θ = σ × Irr rv ( M N ) . Note that o ( θ ) = 1 and 2 θ ( 1 ) . By Lemma 2.7, there is χ Irr rv ( G θ ) , and this is impossible since χ ( 1 ) is not a composite number.

Conclusion: We have proved, so far, that S G Aut ( S ) and that 𝑆 is isomorphic to either A 5 , A 6 or PSL 2 ( 8 ) . Now, 𝑆 cannot be the alternating group A 6 because each of the 5 subgroups between 𝑆 and Aut ( S ) has a rational irreducible character of degree 10 (it is possible to check this with the software GAP), so S { A 5 , PSL 2 ( 8 ) } . In any of these cases, [ Aut ( S ) : S ] is a prime number and there is only one subgroup strictly above 𝑆, namely Aut ( S ) itself. But both Aut ( A 5 ) and Aut ( PSL 2 ( 8 ) ) have a real irreducible character with composite degree. Hence G = A 5 or G = PSL 2 ( 8 ) . ∎

Proposition 3.2

Let 𝐺 be a finite non-solvable group such that cd rv ( G ) consists of prime-power numbers. Then G = K R with R = Rad ( G ) and K = G ( ) . Moreover, K R = L is a 2-group, and K / L is isomorphic to A 5 or PSL 2 ( 8 ) .

## Proof

Let K = G ( ) be the last term of the derived series of 𝐺, and define G ¯ = G / K R . Observe that quotients preserve the hypotheses. Hence, by Theorem 3.1, G / R is a simple group. Since 1 < K R / R G / R , we have that G = K R and K ¯ G / R is isomorphic to A 5 or PSL 2 ( 8 ) . Moreover, G ¯ = K ¯ × R ¯ because [ K , R ] L .

Suppose by contradiction that there is some θ Irr rv ( R ¯ ) of non-trivial degree. By Theorems 2.4 and 2.2, there are two non-linear characters ϕ 1 , ϕ 2 Irr rv ( K ¯ ) such that ϕ 1 ( 1 ) is even and ϕ 2 ( 1 ) is odd. If θ ( 1 ) is odd, consider χ = θ ϕ 1 , and if θ ( 1 ) is even, consider χ = θ ϕ 2 . In any case, 𝜒 is a composite number, but this is impossible. It follows that every real character of R / L is linear, and by Theorem 2.1, R ¯ = O ¯ × H ¯ , where O Hall 2 ( R ) and H Syl 2 ( R ) . Write G 0 for the preimage in 𝐺 of K ¯ H ¯ ; note that G 0 is a normal subgroup of odd index in 𝐺. Note that G 0 = L K H = K H . By Lemma 2.6, cd rv ( G 0 ) consists of prime-power numbers. Moreover, K = G 0 ( ) and Rad ( G 0 ) K = L . Hence we can assume that G = G 0 . This implies that O L .

Suppose, working by contradiction, that O > 1 , namely 𝐿 is not a 2-group. Consider M / M 0 , the first term (from above) of a principal series of 𝐺 such that M , M 0 L and M / M 0 is not a 2-group. Hence M / M 0 is an elementary abelian 𝑝-group for 𝑝 odd, and L / M is a 2-group. Possibly replacing 𝐺 with G / M 0 , we can assume that M 0 = 1 and 𝑀 is a minimal normal subgroup of 𝐺.

Since K / L is simple, C K ( M ) = K or C K ( M ) L . If C K ( M ) = K , then 𝑀 has a direct complement 𝑁 in 𝐿, and we consider K ¯ = K / N . Note that

1 < M ¯ Z ( K ¯ ) K ¯

since K = K is perfect, and hence | M | divides | M ( G ) | by [6, Corollary 11.20], where M ( G ) denotes the Schur multiplier of 𝐺. But this is impossible since | M ( A 5 ) | = 2 and M ( PSL 2 ( 8 ) ) = 1 .

Hence C K ( M ) L , and the action of 𝐾 on 𝑀 is non-trivial. Moreover, K / L has even order, so by Lemma 2.9, there are an element λ M ^ and x K such that λ x = λ ¯ . Let I = I G ( λ ) , and note that x N G ( I ) I , so 2 divides [ G : I ] .

Let I ¯ = I / ker ( λ ) (we remark that the “bar” notation here is not the same as in first part of the proof), and observe that M ¯ Z ( I ¯ ) . Take P Syl p ( I ) ; since the index of 𝐾 in 𝐺 is a 2-power, every subgroup of 𝐺 with odd order is contained in 𝐾; hence P K . Moreover, M ¯ Z ( P ¯ ) , P ¯ Syl p ( I ¯ ) , and P L / L is a 𝑝-subgroup of the simple group K / L that is isomorphic to A 5 or PSL 2 ( 8 ) . Now, if 𝑝 is an odd prime, every Sylow 𝑝-subgroup of A 5 or PSL 2 ( 8 ) is cyclic (see Tables 1 and 2). Hence P / M P ¯ / M ¯ P L / L is cyclic and P ¯ is abelian.

Since M ¯ Z ( I ¯ ) , we have that M ¯ I ¯ by [8, Theorem 5.3]. In addition, we have M ¯ I ¯ = 1 because M ¯ has order 𝑝. Write

I ¯ / I ¯ = Q × B , where B Hall p ( I ¯ / I ¯ ) and Q Syl p ( I ¯ / I ¯ ) .

Note that 𝑄 and 𝐵 are 𝑥-invariant as 𝑥 normalizes 𝐼. By abuse of notation, we write M Q in the place of M ¯ I ¯ / I ¯ Q . In this notation, 𝑀 is a group of order 𝑝, and 𝜆 is a faithful character of 𝑀. The 2-group x acts on the abelian group 𝑄; hence by Maschke’s Theorem [10, Theorem 8.4.6], there is an x -invariant complement 𝑇 for the x -invariant subgroup 𝑀, so Q = M × T . Letting λ ^ = λ × 1 T Irr ( Q ) and δ = λ ^ × 1 B Irr ( I ¯ / I ¯ ) , we have that

δ x = λ ^ x × 1 B x = ( λ x × 1 T x ) × 1 B = ( λ ¯ × 1 T ) × 1 B = δ ¯ .

We return to the previous notation, so 𝛿 lifts to a character of 𝐼 that we also call 𝛿. Note that I < G as 2 divides [ G : I ] .

If I H < G , then I H / H is a proper subgroup of G / H that is a simple group isomorphic to A 5 or PSL 2 ( 8 ) . The maximal subgroups of these two groups are known as well as their indexes; see Tables 1 and 2. In particular, there always is an odd prime 𝑞 such that 𝑞 divides [ G : I H ] , and hence 2 q divides [ G : I ] . Note that δ Irr ( I λ ) , so χ = δ G Irr ( G ) . Moreover,

χ ¯ = ( δ ¯ ) G = ( δ x ) G = δ G = χ .

Hence 𝜒 is a real character of 𝐺, and 2 q χ ( 1 ) since 2 q [ G : I ] , and this is impossible.

Suppose now I H = G . In this case, I / I H G / H , which is isomorphic to A 5 or PSL 2 ( 8 ) . These groups have a unique rational character 𝜙 of odd degree. The element 𝑥 stabilizes the section I / I H ; hence by uniqueness, ϕ x = ϕ . By Gallagher’s Theorem [6, Theorem 6.17], ϕ δ Irr ( I λ ) , and by the Clifford correspondence, χ = ( ϕ δ ) G Irr ( G ) . Since 𝜙 is a real 𝑥-invariant character and δ x = δ ¯ , we have that ( ϕ δ ) x = ϕ δ ¯ . Hence, as before, 𝜒 is a real irreducible character. Now θ ( 1 ) χ ( 1 ) , and there is an odd prime 𝑞 such that 𝑞 divides χ ( 1 ) . Moreover, 2 χ ( 1 ) since 2 divides [ G : I ] . So χ ( 1 ) is a composite number, and this is impossible. ∎

We give the list of maximal subgroups of A 5 and PSL 2 ( 8 ) and their indices.

### Table 1

Maximal subgroups of A 5 .

A 4 D 10 S 3
12 10 6
5 6 10

### Table 2

Maximal subgroups of PSL 2 ( 8 ) .

F 56 D 18 D 14
56 18 14
9 28 72
Lemma 3.3

Let 𝐾 be a perfect group and 𝑀 a minimal normal subgroup of 𝐾 that is an elementary abelian 2-group. Suppose that 𝑀 is non-central in 𝐾 and K / M is isomorphic to L 2 ( 8 ) or A 5 . Then 𝐾 has an irreducible non-linear real character with odd composite degree.

## Proof

Since G / M is simple, we have that C G ( M ) = M . Suppose that K / M is isomorphic to A 5 . There are two non-isomorphic irreducible A 5 -modules W 1 , W 2 of A 5 over GF ( 2 ) . Both have dimension 4, and H 2 ( A 5 , W 1 ) = H 2 ( A 5 , W 2 ) = 0 . Hence 𝑀 has a complement 𝑆 in 𝐾. It is easy to construct these groups, and we see that K = M S = W i A 5 has a real irreducible character of degree 15. Suppose now that K / M L 2 ( 8 ) . Let W 1 , W 2 , W 3 be the non-isomorphic irreducible L 2 ( 8 ) -modules over GF ( 2 ) , where dim ( W 1 ) = 6 , dim ( W 2 ) = 8 , dim ( W 3 ) = 12 . If M M i with i = 2 , 3 , then H 2 ( L 2 ( 8 ) , W i ) = 0 , and hence M i has a complement 𝑆 in 𝐾. Then, as before, we conclude by observing that W i L 2 ( 8 ) has a real irreducible character of degree 63. Suppose that M W 1 . Then we have dim H 2 ( L 2 ( 8 ) , W 1 ) = 3 . Nevertheless, there are just two perfect groups of order 2 6 | L 2 ( 8 ) | . Both these groups have an irreducible real character of degree 63. ∎

In the previous lemma, dimensions of cohomology groups and all the perfect groups of a given order is information that is accessible with the GAP functions cohomolo and PerfectGroup.

Proposition 3.4

Let 𝐺 be a finite non-solvable group, and suppose that cd rv ( G ) consists of prime-power numbers. Let K = G ( ) and R = Rad ( G ) . Then we have | K R | 2 , and if equality holds, then K SL 2 ( 5 ) .

## Proof

By Proposition 3.2, we have that N = K R is a 2-group. We prove that if N > 1 , then | N | = 2 and 𝐾 is isomorphic to SL 2 ( 5 ) . Let V = N / Φ ( N ) ; then 𝑉 is a normal elementary abelian 2-subgroup of G / Φ ( N ) . Let V > V 1 > > V n be a 𝐾-principal series of 𝑉. Let N > N 1 > > N n be such that N i is the preimage in 𝑁 of V i . Then N / N 1 is an irreducible K / N -module, and K / N is isomorphic A 5 or L 2 ( 8 ) by Proposition 3.2. By Lemmas 3.3 and 2.7, N / N 1 is central in K / N 1 . Since 𝐾 is perfect, we have that N / N 1 is isomorphic to a subgroup of the Schur multiplier M ( K / N ) . The only possibility is | N / N 1 | = 2 and K / N 1 SL 2 ( 5 ) , the Schur covering of A 5 . Suppose by contradiction that N 1 / N 2 > 1 ; write K ¯ = K / N 2 . Since M ( SL 2 ( 5 ) ) = 1 , N ¯ 1 cannot be central in K ¯ . Let t K be a 2-element such that t N 1 = Z ( K / N 1 ) , namely the unique central involution in SL 2 ( 5 ) and t N 1 = O 2 ( K / N 1 ) . Since N 1 is an irreducible module over GF ( 2 ) , we have that 𝑡 acts trivially on N ¯ 1 . Suppose that t ¯ 2 1 ; then t ¯ 2 would be a proper, non-trivial submodule of N ¯ 1 , against irreducibility. This means that t ¯ 2 = 1 and hence t ¯ , which centralizes N ¯ 2 , is a minimal normal subgroup of N ¯ 2 . Observe that K ¯ / t is a quotient of 𝐾 that satisfies the hypotheses of Lemma 3.3. Hence, by Lemma 2.7, we derive a contradiction. ∎

We now prove Theorem A, which we restate for convenience of the reader.

Theorem 3.5

Let 𝐺 be a finite non-solvable group, and suppose that cd rv ( G ) consists of prime-power numbers. Then Rad ( G ) = H × O for a group 𝑂 of odd order and a 2-group 𝐻 of Chillag–Mann type. Furthermore, if K = G ( ) , then one of the following holds:

1. G = K × R , and 𝐾 is isomorphic to A 5 or L 2 ( 8 ) ;

2. G = ( K H ) × O , where K SL 2 ( 5 ) , H K = H \Ydown K , K H = Z ( K ) < H .

## Proof

By Proposition 3.4 and Proposition 3.2, if K = G ( ) and R = Rad ( G ) , then G = K R , and either K R = 1 and 𝐾 is simple isomorphic to A 5 or L 2 ( 8 ) or K SL 2 ( 5 ) and K R = Z ( K ) . In the first case, point (i) follows. Suppose K = SL 2 ( 5 ) and K R = Z ( K ) = Z . Note that 𝑍 is a normal subgroup of order 2, hence is central in 𝑅. Consider G ¯ = G / Z . Then G ¯ = K ¯ × R ¯ , and hence R ¯ is a group of Chillag–Mann type since K ¯ is simple and has irreducible real non-linear characters of both odd and even degree. This means that R ¯ = O ¯ × H ¯ for O Hall 2 ( R ) and H Syl 2 ( R ) ; note that H ¯ is of Chillag–Mann type. We have that 𝑅 is 2-closed, and hence R = H O . Clearly, 𝑂 acts trivially on H / Z , so

H = C H ( O ) Z C H ( O ) Z ( R ) H .

It follows that 𝑂 centralizes 𝐻 and R = H × O . By the Dedekind modular law, H K O H K R H ( K R ) H , and hence H K O H O = 1 . This means that G = ( K H ) × O and K H = Z , which has order 2. If 𝐻 and 𝐾 commute, then K H = K \Ydown H . Suppose by contradiction that [ H , K ] = Z ; hence there is a Z H / Z that acts non-trivially by conjugation on K / Z . But this is impossible since K H / Z = K ¯ × H ¯ . We now prove that 𝐻 is of Chillag–Mann type. Suppose that this is not the case, so there is ϕ Irr rv ( H ) such that ϕ ( 1 ) > 1 . Since H ¯ is of Chillag–Mann type, we have that Z ker ϕ , so ϕ Z = ϕ ( 1 ) λ , with λ 1 Z . On the other hand, if 𝜃 is the unique character of 𝐾 of degree 6, then Z ker θ and θ Z = θ ( 1 ) λ . Now, K H = K × H / N , where N = { ( z , z ) z Z } (see [5, I9.10]) and ψ = θ × ϕ Irr rv ( K × H ) . Moreover,

ψ N = ϕ ( 1 ) θ ( 1 ) λ 2 = ϕ ( 1 ) θ ( 1 ) 1 N ,

so it follows that N ker ψ and ψ Irr rv ( K H ) . If χ = ψ × 1 O , then χ Irr ( G ) takes real values and has composite degree, which is impossible. Since SL 2 ( 5 ) does not satisfy the hypotheses, we have that Z < H . Point (ii) follows. ∎

We remark that, in [6, Problem 4.4], we can find a stronger version of the argument used in the proof above.

As a consequence, we get Theorem B.

Corollary 3.6

Let 𝐺 be a non-solvable group, and suppose that cd rv ( G ) consists of prime-power numbers. Then either

cd rv ( G ) = cd rv ( L 2 ( 8 ) ) or cd rv , 2 ( G ) = cd rv , 2 ( A 5 ) .

## Proof

Apply Theorem 3.5. In case (i), there is nothing to prove. Suppose (ii). Then we have that G = ( K H ) × O with 𝑂 of odd order, K = G ( ) and 𝐻 a normal 2-subgroup. Let 𝑆 denote the simple section K H / H ; hence S A 5 . Take χ Irr rv ( G ) to be a real non-linear character of odd degree. Hence χ ( 1 ) = p n with 𝑝 odd and 𝜒 is a character of H K since, by Lemma 2.5, O ker ( χ ) . The degree of every irreducible constituent of χ H divides ( | H | , χ ( 1 ) ) = 1 , and hence χ H = e i λ i for λ i Lin ( H ) . By hypothesis, we have that χ ( 1 ) = p * > 1 for an odd prime 𝑝, and by [6, Corollary 11.29], χ ( 1 ) / λ ( 1 ) divides [ H K : H ] = | S | , where S A 5 . Hence p χ ( 1 ) | S | p , the 𝑝-part of the number | S | , which is equal to 𝑝 if 𝑝 is an odd prime. It follows that χ ( 1 ) = p . We have proved that cd rv , 2 ( G ) { 3 , 5 } = cd rv , 2 ( A 5 ) . The right-to-left inclusion follows by observing that A 5 is a quotient of 𝐺. ∎

# Acknowledgements

The author would like to thank Professor Silvio Dolfi for his suggestions and comments to the first manuscript.

1. Communicated by: Britta Späth

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